I think this may be what you are trying to do. where the n can be any number.

f[n_] := Cases[
  Tuples[Range[n], {3}], {a_, b_, c_} /; 
   a <= b <= c && c <= a + b && IntegerQ[Sqrt[2 a^2 + 2 b^2 - c^2]/2]]


 f[10]

 {{1, 1, 2}, {1, 3, 4}, {1, 5, 6}, {1, 7, 8}, {1, 9, 10}, {2, 
  2, 4}, {2, 4, 6}, {2, 6, 8}, {2, 8, 10}, {3, 3, 6}, {3, 5, 8}, {3, 
  7, 10}, {4, 4, 8}, {4, 6, 10}, {5, 5, 6}, {5, 5, 8}, {5, 5, 10}, {6,
   8, 10}}

Paul.

I noticed that some of those results were zero, so if we want to test for the Integer part to be greater than zero we can add another test like so

f[n_] := Cases[
  Tuples[Range[n], {3}], {a_, b_, c_} /; 
   a <= b <= c && c <= a + b && c > 0 && 
    IntegerQ[p = Sqrt[2 a^2 + 2 b^2 - c^2]/2] && p > 0]

Notice I set the sqrt part to equal p and then tested the p to be greater than zero afterwards.

P

Thank you . I do not master properly the language. In your code ,I tried to change # + #2 into #1 +#2 and that changes nothing, (??) The initial condition a ? b ? c < a+b is not fullfilled by the function LessEqual but if this one is replaced by Less, we loose all cases where a = b or b = c. What I absolutely do not see is the way your code gives a ? b ? c, because the function which operates seems to give only that every composant is less than the sum of the two others. (# ? #1 + #2) In my diophantine problem,I have in fact c necessarily even. I cannot add this condition because the third composant c is not identified with the symbol #. Do not waste your patience if my questions are childish. Thank you for your help if you take time to teach.

Best regards

The only question that puzzles me is that I may read the code erroneously because the condition a ? b ? c is not visible ( I understand that the code gives a ? b+c, b ? c+a , c ? a+b ). This reading must be wrong.

Semi automatic, brute force approach, but works for less memory consuming examples:

Select[Tuples[Range[10], {3}], # <= #2 <= #3 < # + #2 & @@ # &   ]